裸kmp算法

Number Sequence

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

 

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

 

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

 

Sample Input

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

 

Sample Output

6

-1

 

刚接触kmp还不是很理解，next失效函数

 

 #include <iostream>
 #include <cstdio>
 using namespace std;
 int f[];
 int n,m;
 int a[],b[];
 int kmp(int t[],int p[])
 {
     int i=,j=;
     while(i<n&&j<m)
     {
         if(j==-||t[i]==p[j])
         {
             i++;
             j++;
         }
         else
             j=f[j];
     }
     if(j>=m)
         return i-m+;
     return -;
 }
 void next(int p[])
 {
     int k=-,j=;
     f[]=-;
     while(j<m)
     {
         if(k==-||p[k]==p[j])
         {
             k++;
             f[++j]=k;
         }
         else
             k=f[k];
     }
 }
 int main()
 {
     int T,i,j;
     //freopen("in.txt","r",stdin);
     cin>>T;
     while(T--)
     {
         cin>>n>>m;
         for(i=;i<n;i++)
             cin>>a[i];
         for(i=;i<m;i++)
             cin>>b[i];
         next(b);
         cout<<kmp(a,b)<<endl;
     }
 }